From: Jens Kruse Andersen
Date: 12/31/05 12:02:34
To: Prime Numbers
Subject: Re: [PrimeNumbers] Prime GAP of 364,188
=20
Jose Ramòn Brox wrote:
> > There are better methods to choose which numbers have small factors.
> > Such methods were used for the 2 known Megagaps:
> >
http://hjem.get2net.dk/jka/math/primegaps/megagap.htm
>
> Indeed a very intelligent approach: I was wondering how all that records
> could be arrived at, and now I know. Are there any other approaches,
> or yours is the main used one?
AFAIK, all efficient searches for large gaps between large primes/prp's (e.g
above 50 digits) have included some variant of these points:
********
From Kermit
kermit@...
Has anyone tried this approach?
To product a prime gap
choose prime divisors,
p1, p2, p3, .... pN.
Lets pick p1 = 2, p2 = 3, p3 = 5, etc.
Choose k = -1 mod 2. This makes k odd.
choose k = -2 mod 3.
Now we don't need to worry about -3, -5, -7, etc because these are all taken
care of by choosing k = -1 mod 2.
choose k = - 4 mod 5.
Now we don't need to worry about -5, - 8, - 11, - 14, etc, because these are
already taken care of by k = -2 mod 3.
choose k = -6 mod 7.
Now we don't need to worry about -9, -14, -19, - 24, etc because these are
taken care of by k = -4 mod 5.
Choose k = - 10 mod 11.
Now we don't need to worry about -6, -13, -20, etc because these are taken
care of by=20 k = -6 mod 7.
etc.
So with the 5 primes 2,3,5,7,11 we can assure a prime gape of length 11.
Each prime we add to the list lengthens the prime gap by more than 1.
In fact, note that we just defined k = 1 mod 2, k = 1 mod 3, k = 1 mod 5,
k = 1 mod 7.
So the minimum k satisfying this is
k = 2 * 3 * 5 * 7 * 11 + 1 = 2311.
In fact, I just realized why the pattern I chose will always yield p#+1 as
the value of k. Thus I just proved that the prime gap of p# is at least p.
Perhaps some other way of assigning the negative values will produce a sieve
of higher merit.
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